Question

Electromagnetic wave of-intensity 1400 W/m² falls normally on a surface of area 1.5 m² is completely absorbed by it. Find out force exerted by beam.

• A. 14 × 10⁻⁵ N
• B. 14 × 10⁻⁶ N
• C. 7 × 10⁻⁵ N
• D. 7 × 10⁻⁶ N

Answer 1

Answer: (D)

7 × 10⁻⁶ N

Answer 2

The intensity $I$ of an electromagnetic wave is the power per unit area. The total power $P$ incident on a surface of area $A$ is given by $P = IA$. The momentum $p$ carried by electromagnetic radiation of energy $E$ is $p = E/c$, where $c$ is the speed of light. The force exerted by the electromagnetic beam on a surface is the rate of momentum transfer. For a surface that completely absorbs the radiation, the momentum transferred per unit time is equal to the momentum of the incident radiation per unit time. Thus, the force $F$ is given by: $F = \frac{\text{Momentum}}{\text{Time}} = \frac{\text{Energy/Time}}{c} = \frac{P}{c}$ Substituting $P = IA$, we get: $F = \frac{IA}{c}$

Given: Intensity, $I = 1400 \, \text{W/m}^2$ Area, $A = 1.5 \, \text{m}^2$ Speed of light, $c \approx 3 \times 10^8 \, \text{m/s}$

Substituting the values into the formula: $F = \frac{(1400 \, \text{W/m}^2) \times (1.5 \, \text{m}^2)}{3 \times 10^8 \, \text{m/s}}$ $F = \frac{2100}{3 \times 10^8} \, \text{N}$ $F = \frac{700}{10^8} \, \text{N}$ $F = 700 \times 10^{-8} \, \text{N}$ $F = 7 \times 10^{-6} \, \text{N}$